A thin circular ring of area $$A$$ is held perpendicular to a
uniform magnetic field of induction $$B.$$ $$A$$ small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is $$R.$$ When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer is
A.
$$\frac{{BR}}{A}$$
B.
$$\frac{{AB}}{R}$$
C.
$$ABR$$
D.
$$\frac{{{B^2}A}}{{{R^2}}}$$
Answer :
$$\frac{{AB}}{R}$$
Solution :
The current induced will be
$$\eqalign{
& i = \frac{{\left| e \right|}}{R} \Rightarrow i = \frac{1}{R}\frac{{d\phi }}{{dt}}\,\,{\text{But }}i = \frac{{dq}}{{dt}} \cr
& \Rightarrow \quad \frac{{dq}}{{dt}} = \frac{1}{R}\frac{{d\phi }}{{dt}} \Rightarrow \int d q = \frac{1}{R}\int d \phi \Rightarrow q = \frac{{BA}}{R} \cr} $$
Releted MCQ Question on Electrostatics and Magnetism >> Electromagnetic Induction
Releted Question 1
A thin circular ring of area $$A$$ is held perpendicular to a
uniform magnetic field of induction $$B.$$ $$A$$ small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is $$R.$$ When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer is
A thin semi-circular conducting ring of radius $$R$$ is falling with its plane vertical in horizontal magnetic induction $$\overrightarrow B .$$ At the position $$MNQ$$ the speed of the ring is $$v,$$ and the potential difference developed across the ring is
A.
zero
B.
$$\frac{{Bv\pi {R^2}}}{2}$$ and $$M$$ is at higher potential
Two identical circular loops of metal wire are lying on a table without touching each other. Loop-$$A$$ carries a current which increases with time. In response, the loop-$$B$$
A coil of inductance $$8.4 mH$$ and resistance $$6\,\Omega $$ is connected to a $$12 V$$ battery. The current in the coil is $$1.0 A$$ at approximately the time